% © Evan Chen
% Downloaded from https://web.evanchen.cc/
\documentclass[11pt]{scrartcl}
\usepackage[sexy]{evan}
\ihead{\footnotesize\textbf{\thetitle}}
\ohead{\footnotesize\href{http://web.evanchen.cc}{\ttfamily web.evanchen.cc},
updated \today}
\title{IMO 2021 Solution Notes}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
This is a compilation of solutions
for the 2021 IMO.
Some of the solutions are my own work,
but many are from the official solutions provided by the organizers
(for which they hold any copyrights),
and others were found by users on the Art of Problem Solving forums.
These notes will tend to be a bit more advanced and terse than the ``official''
solutions from the organizers.
In particular, if a theorem or technique is not known to beginners
but is still considered ``standard'', then I often prefer to
use this theory anyways, rather than try to work around or conceal it.
For example, in geometry problems I typically use directed angles
without further comment, rather than awkwardly work around configuration issues.
Similarly, sentences like ``let $\mathbb{R}$ denote the set of real numbers''
are typically omitted entirely.
Corrections and comments are welcome!
\end{abstract}
\tableofcontents
\newpage
\addtocounter{section}{-1}
\section{Problems}
\begin{enumerate}[\bfseries 1.]
\item %% Problem 1
Let $n \ge 100$ be an integer.
Ivan writes the numbers $n, n+1, \ldots, 2n$ each on different cards.
He then shuffles these $n+1$ cards, and divides them into two piles.
Prove that at least one of the piles contains two cards such that
the sum of their numbers is a perfect square.
\item %% Problem 2
Show that the inequality
\[\sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i-x_j|}
\le \sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i+x_j|} \]
holds for all real numbers $x_1$, $x_2$, \dots, $x_n$.
\item %% Problem 3
Let $D$ be an interior point of the acute triangle $ABC$
with $AB > AC$ so that $\angle DAB = \angle CAD$.
The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD$,
the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC$,
and the point $X$ on the line $AC$ satisfies $CX = BX$.
Let $O_1$ and $O_2$ be the circumcenters of the triangles
$ADC$ and $EXD$, respectively.
Prove that the lines $BC$, $EF$, and $O_1O_2$ are concurrent.
\item %% Problem 4
Let $\Gamma$ be a circle with center $I$, and $ABCD$ a convex quadrilateral
such that each of the segments $AB$, $BC$, $CD$ and $DA$ is tangent to $\Gamma$.
Let $\Omega$ be the circumcircle of the triangle $AIC$.
The extension of $BA$ beyond $A$ meets $\Omega$ at $X$,
and the extension of $BC$ beyond $C$ meets $\Omega$ at $Z$.
The extensions of $AD$ and $CD$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively.
Prove that
\[ AD + DT + TX + XA = CD + DY + YZ + ZC. \]
\item %% Problem 5
Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter.
Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes
in a circular pattern in the ground around their favourite tree.
The next morning Jumpy notices that Bushy had placed one walnut into each hole,
but had paid no attention to the numbering.
Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves.
In the $k$th move, Jumpy swaps the positions
of the two walnuts adjacent to walnut $k$.
Prove that there exists a value of $k$ such that, on the $k$th move,
Jumpy swaps some walnuts $a$ and $b$ such that $a AC$ so that $\angle DAB = \angle CAD$.
The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD$,
the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC$,
and the point $X$ on the line $AC$ satisfies $CX = BX$.
Let $O_1$ and $O_2$ be the circumcenters of the triangles
$ADC$ and $EXD$, respectively.
Prove that the lines $BC$, $EF$, and $O_1O_2$ are concurrent.
\end{mdframed}
\emph{This problem and solution were contributed by Abdullahil Kafi}.
\begin{claim*}
Quadrilateral $BCEF$ is cyclic.
\end{claim*}
\begin{proof}
Let $D'$ be the isogonal conjugate of the point $D$. The
angle condition implies quadrilateral $CEDD'$ and $BFDD'$
are cyclic. By power of point we have \[ AE\cdot AC=AD\cdot AD'=AF\cdot AB \]
So $BCEF$ is cyclic.
\end{proof}
\begin{claim*}
Line $ZD$ is tangent to the circles $(BCD)$ and $(DEF)$
where $Z=EF\cap BC$.
\end{claim*}
\begin{proof}
Let $\angle CAD=\angle BAD=\alpha$, $\angle BCD=\beta$,
$\angle DBC=\gamma$, $\angle ACD=\phi$,
$\angle ABD=\epsilon$.
From $\triangle ABC$ we have
$2\alpha+\beta+\gamma+\phi+\epsilon=180^\circ$.
Let $\ell$ be a line tangent to $(BCD)$ and $K$ be a
point on it in the same side of $AD$ as $C$ and
$L=AD\cap BC$. From our labeling we have,
\begin{align*}
\angle AFE &= \beta + \phi \qquad \angle BFD =
\alpha + \gamma \qquad \angle DFE = \alpha + \phi
\qquad \angle CDL = \alpha + \phi
\end{align*}
Now $\angle CDJ = 180^\circ - \gamma - \beta - (\alpha + \phi) = \alpha + \epsilon$.
So $\angle DFE = \angle EDK = \alpha + \epsilon$, which
means $\ell$ is also tangent to $(DEF)$. Now by the
radical center theorem we have $\ell$ passes through
$Z$.
\end{proof}
Let $M$ be the Miquel point of the cyclic quadrilateral
$BCEF$. From the Miquel configuration we have $A$, $M$, $Z$
are collinear and $(AFEM)$, $(ZCEM)$ are cyclic.
\begin{claim*}
Points $B$, $X$, $M$, $E$ are cyclic.
\end{claim*}
\begin{proof}
Notice that $\angle EMB = 180^\circ - \angle AMB -\angle EMZ$
$=$ $180^\circ - 2\angle ACB = \angle EXB$.
\end{proof}
Let $N$ be the other intersection of circles $(ACD)$ and
$(DEX)$ and let $R$ be the intersection of $AC$ and $BM$.
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
import graph;
size(14cm);
pen ffxfqq = rgb(1.,0.49803,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen ffdxqq = rgb(1.,0.84313,0.);
pen qqffff = rgb(0.,1.,1.);
draw((12.,-7.5)--(3.88816,0.36923), linewidth(0.4) + red);
draw((3.88816,0.36923)--(0.5,-7.5), linewidth(0.4) + red);
draw((2.73129,-2.31767)--(4.74594,-3.92841), linewidth(0.4));
draw((4.74594,-3.92841)--(5.47993,-1.17493), linewidth(0.4));
draw(circle((6.25,-6.90419), 5.78078), linewidth(0.4) + red);
draw((6.25,5.85474)--(12.,-7.5), linewidth(0.4));
draw((-9.73375,-7.5)--(5.47993,-1.17493), linewidth(0.4) + red);
draw((-9.73375,-7.5)--(12.,-7.5), linewidth(0.4) + red);
draw((-9.73375,-7.5)--(4.74594,-3.92841), linewidth(0.4));
draw(circle((0.03259,-2.63473), 4.88766), linewidth(0.4) + ffxfqq);
draw(circle((6.84056,0.75675), 5.13208), linewidth(0.4) + qqwuqq);
draw(circle((13.09242,3.87122), 11.42357), linewidth(0.4) + ffxfqq);
draw(circle((6.25,-5.31169), 6.15233), linewidth(0.4) + red);
draw(circle((3.88816,-1.22327), 1.59250), linewidth(0.4) + red);
draw((3.88816,0.36923)--(-9.73375,-7.5), linewidth(0.4));
draw(circle((9.81689,-0.52472), 7.30892), linewidth(0.4) + ffdxqq);
draw((2.50861,-0.42772)--(12.,-7.5), linewidth(0.4));
draw((1.83901,1.90685)--(4.74594,-3.92841), linewidth(0.4));
draw((6.25,5.85474)--(3.88816,0.36923), linewidth(0.4));
draw((0.5,-7.5)--(4.74594,-3.92841), linewidth(0.4));
draw((4.74594,-3.92841)--(12.,-7.5), linewidth(0.4));
draw(shift((-9.73375,-7.5))*xscale(14.91368)*yscale(14.91368)*arc((0,0),1,3.18577,50.41705), linewidth(0.4) + dotted + qqffff);
dot((12.,-7.5),linewidth(3.pt));
label("$B$", (12.47142,-8.47771), NE);
dot((0.5,-7.5),linewidth(3.pt));
label("$C$", (-0.29392,-8.47771), NE);
dot((3.88816,0.36923),linewidth(3.pt));
label("$A$", (3.42645,1.11336), NE);
dot((6.25,5.85474),linewidth(3.pt));
label("$X$", (5.78156,6.47208), NE);
dot((4.74594,-3.92841),linewidth(3.pt));
label("$D$", (4.65520,-5.03038), NE);
dot((5.47993,-1.17493),linewidth(3.pt));
label("$F$", (5.64503,-0.86628), NE);
dot((2.73129,-2.31767),linewidth(3.pt));
label("$E$", (1.58333,-2.43635), NE);
dot((-9.73375,-7.5),linewidth(3.pt));
label("$Z$", (-10.77243,-8.20465), NE);
dot((2.50861,-0.42772),linewidth(3.pt));
label("$M$", (1.20788,-0.21777), NE);
dot((1.83901,1.90685),linewidth(3.pt));
label("$N$", (0.79829,2.47864), NE);
dot((3.29328,-1.01240),linewidth(3.pt));
label("$R$", (3.63125,-1.10520), NE);
\end{asy}
\end{center}
\begin{claim*}
Points $B$, $D$, $M$, $N$ are cyclic.
\end{claim*}
\begin{proof}
By power of point we have
\[
\opname{Pow}(R, (ACD)) = RC \cdot RA = RM \cdot RB
= RE \cdot RX = \opname{Pow}(R, (DEX)).
\]
Hence $R$ lies on the radical axis of $(ACD)$ and
$(DEX)$, so $N$, $R$, $D$ are collinear. Also
\[ RN \cdot RD = RA \cdot RC = RM \cdot RB \] So $BDMN$
is cyclic.
\end{proof}
Notice that $(ACD)$, $(BDMN)$, $(DEX)$ are coaxial so their
centers are collinear. Now we just need to prove the
centers of $(ACD)$, $(BDMN)$ and $Z$ are collinear. To
prove this, take a circle $\omega$ with radius $ZD$
centered at $Z$. Notice that by power of point
\[ ZC \cdot ZB = ZD^2 = ZE \cdot ZF = ZM \cdot ZA \]
which means inversion circle $\omega$ swaps $(ACD)$ and $(BDMN)$.
So the centers of $(ACD)$ and $(BDMN)$ must
have to be collinear with the center of inversion circle, as desired.
\pagebreak
\section{Solutions to Day 2}
\subsection{IMO 2021/4, proposed by Dominik Burek (POL) and Tomasz Ciesla (POL)}
\textsl{Available online at \url{https://aops.com/community/p22698001}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $\Gamma$ be a circle with center $I$, and $ABCD$ a convex quadrilateral
such that each of the segments $AB$, $BC$, $CD$ and $DA$ is tangent to $\Gamma$.
Let $\Omega$ be the circumcircle of the triangle $AIC$.
The extension of $BA$ beyond $A$ meets $\Omega$ at $X$,
and the extension of $BC$ beyond $C$ meets $\Omega$ at $Z$.
The extensions of $AD$ and $CD$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively.
Prove that
\[ AD + DT + TX + XA = CD + DY + YZ + ZC. \]
\end{mdframed}
Let $PQRS$ be the contact points of $\Gamma$ an $\ol{AB}$, $\ol{BC}$,
$\ol{CD}$, $\ol{DA}$.
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair I = (0.,0.);
pair P = (-0.40442,0.91457);
pair Q = (-0.35670,-0.93421);
pair R = (0.99998,0.00468);
pair S = (0.70710,0.70710);
pair A = (0.22244,1.19177);
pair B = (-2.62593,-0.06777);
pair C = (1.00681,-1.45484);
pair D = (0.99806,0.41615);
pair X = (2.38106,2.14630);
pair Z = (1.47042,-1.63185);
pair Y = (2.86072,-1.44650);
pair T = (0.99083,1.96044);
pair E = (0.99283,1.53243);
pair F = (4.01929,-2.60507);
size(12.41979cm);
pen qqwuqq = rgb(0.,0.39215,0.);
pen fuqqzz = rgb(0.95686,0.,0.6);
pen zzttqq = rgb(0.6,0.2,0.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(P--Q--R--S--cycle, linewidth(0.6) + zzttqq);
draw(circle(I, 1.), linewidth(0.6) + qqwuqq);
draw(circle((1.92609,0.25714), 1.94318), linewidth(0.6) + fuqqzz);
draw(P--Q, linewidth(0.6) + zzttqq);
draw(Q--R, linewidth(0.6) + zzttqq);
draw(R--S, linewidth(0.6) + zzttqq);
draw(S--P, linewidth(0.6) + zzttqq);
draw(circle((0.50340,-0.72742), 0.88462), linewidth(0.6) + fuqqzz);
draw(circumcircle(I,P,S), dotted + fuqqzz);
draw(A--F, linewidth(0.6) + qqwuqq);
draw(B--X, linewidth(0.6) + qqwuqq);
draw(B--F, linewidth(0.6) + qqwuqq);
draw(C--T, linewidth(0.6) + qqwuqq);
dot("$I$", I, dir(160));
dot("$P$", P, dir((-17.392, 6.881)));
dot("$Q$", Q, dir((-17.176, -15.513)));
dot("$R$", R, dir((2.333, 5.318)));
dot("$S$", S, dir((2.248, 5.460)));
dot("$A$", A, dir((-11.357, 7.426)));
dot("$B$", B, dir((-5.839, 9.792)));
dot("$C$", C, dir((2.205, -22.196)));
dot("$D$", D, dir(45));
dot("$X$", X, dir(80));
dot("$Z$", Z, dir((-3.145, -21.676)));
dot("$Y$", Y, dir(90));
dot("$T$", T, dir((-10.053, 11.473)));
dot("$E$", E, dir((10.253, -19.990)));
dot("$F$", F, dir((2.446, 4.708)));
\end{asy}
\end{center}
\begin{claim*}
We have $\triangle IQZ \cong \triangle IRT$.
Similarly, $\triangle IPX \cong \triangle ISY$.
\end{claim*}
\begin{proof}
By considering $(CQIR)$ and $(CITZ)$,
there is a spiral similarity similarity
mapping $\triangle IQZ$ to $\triangle IRT$.
Since $IQ = IR$, it is in fact a congruence.
\end{proof}
This congruence essentially solves the problem.
First, it implies:
\begin{claim*}
$TX = YZ$.
\end{claim*}
\begin{proof}
Because we saw $IX = IY$ and $IT = IZ$.
\end{proof}
Then, we can compute
\begin{align*}
AD + DT + XA
&= AD + (RT - RD) + (XP-AP) \\
&= (AD-RD-AP) + RT + XP = RT + XP
\end{align*}
and
\begin{align*}
CD + DY + ZC &= CD + (SY-SD) + (ZQ-QC) \\
&= (CD-SD-QC) + SY + ZQ = SY + ZQ
\end{align*}
but $ZQ = RT$ and $XP = SY$, as needed.
\pagebreak
\subsection{IMO 2021/5, proposed by Spain}
\textsl{Available online at \url{https://aops.com/community/p22697921}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter.
Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes
in a circular pattern in the ground around their favourite tree.
The next morning Jumpy notices that Bushy had placed one walnut into each hole,
but had paid no attention to the numbering.
Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves.
In the $k$th move, Jumpy swaps the positions
of the two walnuts adjacent to walnut $k$.
Prove that there exists a value of $k$ such that, on the $k$th move,
Jumpy swaps some walnuts $a$ and $b$ such that $a 0$.
\end{remark*}
\pagebreak
\end{document}